Solution

First the Data

Beam radius: $a = 10\ \text{cm} = 0.1\ \text{m}$.

Wavelength: $\lambda = 10^{-5}\ \text{cm} = 10^{-7}\ \text{m}$

use visible light
$\lambda \approx 5\times10^{-5}\ \text{cm} = 5\times10^{-7}\ \text{m}$

We will adopt $\lambda = 5\times10^{-7}\ \text{m}$.
· Earth–Moon distance:
$L \approx 3.84\times10^{8}\ \text{m}$.

The laser beam emerges from a circular aperture of radius a. Diffraction produces an Airy pattern on the Moon. The first intensity minimum (where the waves begin to cancel) defines the limit of the central spot. For a circular aperture, the angle of the first minimum is:

$\theta \approx \frac{0.61\lambda}{a}$

$R = L\,\theta = \frac{0.61\,\lambda L}{a}$.

Multiply $\lambda L$:

$\lambda L = (5\times10^{-7})\,(3.84\times10^{8}) = 192\ \text{m}$

Divide by a:

$\frac{\lambda L}{a} = \frac{192}{0.1} = 1920\ \text{m}$.

Apply the factor 0.61:

$R = 0.61 \times 1920 = 1171.2\ \text{m}$.

Rounding to order of magnitude:

$\boxed{R \approx 1\ \text{km}}$.

b) Radar antenna of 3 cm

Data

Wavelength: $\lambda = 3\ \text{cm} = 0.03\ \text{m}$.
· Total angular divergence of the beam: $\theta \approx 0.01\ \text{rad}$.

A circular parabolic antenna of diameter D produces a beam whose angular width is limited by diffraction. The angular radius of the first minimum is:

$\theta \approx \frac{\lambda}{D}$.

Solve for D:

$D \approx \frac{\lambda}{\theta} = \frac{0.03}{0.01} = 3\ \text{m}$.

The radius of the antenna is half the diameter:

$R = \frac{D}{2} = 1.5\ \text{m}$.

$\boxed{R \approx 1.5\ \text{m}}$.